1 Chapter 2Energy and Matter 2.6 Changes of State Copyright © 2009 by Pearson Education, Inc.
Chapter 2Energy and Matter 2.4 Specific Heat Copyright © 2009 by Pearson Education, Inc.
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Transcript of Chapter 2Energy and Matter 2.4 Specific Heat Copyright © 2009 by Pearson Education, Inc.
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Chapter 2 Energy and Matter
2.4 Specific Heat
Copyright © 2009 by Pearson Education, Inc.
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Specific heat
• is different for different substances.
• is the amount of heat that raises the temperature of 1 g of a substance by 1 °C.
• in the SI system has units of J/g °C.
• in the metric system has units of cal/g °C.
Specific Heat
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Examples of Specific Heats
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A. When ocean water cools, the surrounding air 1) cools. 2) warms. 3) stays the same.
B. Sand in the desert is hot in the day, and cool at night. Sand must have a
1) high specific heat. 2) low specific heat.
Learning Check
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A. When ocean water cools, the surrounding air
2) warms.(heat is transferred to the air above it. Conservation of Energy)
B. Sand in the desert is hot in the day, and cool
at night. Sand must have a
2) low specific heat.(Low specific heat means that there can be a big temperature increase
with input of a given amount of energy. High specific heat means that there will be a low temperature increase with the input of the same amount of energy. Specific heat is related to “heat capacity”)
Solution
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Learning Check
What is the specific heat if 24.8 g of a metal
absorbs 275 J of energy and the temperature rises
from 20.2 °C to 24.5 °C?
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Solution
Given: 24.8 g metal, 275 J of energy, 20.2 °C to 24.5 °C
Need: Specific heat J/g °C
Plan: specific heat (SH) = Heat (J) g °C
ΔT = 24.5 C – 20.2 °C = 4.3 °C
Setup: 275 J = 2.6 J/g °C (2 sig figs!) (24.8 g)(4.3 °C)
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Heat Equation
The amount of heat lost or gained by a substance iscalculated from the• mass of substance (g).• temperature change (°T).• specific heat of the substance (J/g °C).
This is expressed as the heat equation.
Heat = g x °C x J = J g °C
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How many kJ are needed to raise the temperature of 325 g of water from 15.0 °C to 77.0 °C?
1) 20.4 kJ
2) 77.7 kJ
3) 84.3 kJ
Learning Check
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Study Tip: Using Specific Heat
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Solution
Answer: 3) 84.3 kJ
77.0 °C – 15.0 °C = 62.0 °C
325 g x 62.0 °C x 4.184 J x 1 kJ
g °C 1000 J
= 84.3 kJ
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Coral• contains algae that produce sugars (food) and the
bright red and orange pigments of coral. • expels the algae when water temperatures increase
as little as 1 °C.• bleaches as it loses its food supply and color. • dies if the stress of higher temperatures continues. • reefs in Australia and the Indian Ocean have been
badly damaged by increases in ocean temperatures.
Coral Bleaching
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How many kcal are absorbed by ocean water if 3 x 1018 L of water in the Caribbean has an increase of 1 °C. Assume the specific heat of ocean water is the same as water. Assume the density of ocean water is 1.0 g/mL.
1) 3 x 1015 kcal
2) 3 x 1018 kcal
3) 3 x 1021 kcal
Learning Check
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Solution
183 10 L1000 mL
1 L
1 g
1 mL
21
21
= 3 10 g of sea water
3 10 g
1 °C1 cal
g °C
1 kcal
1000 cal
18
= 3 10 kcal of heat absorbed (Answer 2)
seawaterseawater